Q. 18 - Calculus

Question

Using polar coordinates to evaluate iterated integrals: Sketch the region determined by the limits of the given iterated integrals, and then evaluate the integrals. 

${\int }_{\mathrm{\pi }/2}^{3\mathrm{\pi }/2}{\int }_1^4{r}^{3/2} drd\theta$

Step-by-Step Solution

Answer

${\int }_{\mathrm{\pi }/2}^{3\mathrm{\pi }/2}{\int }_1^4{r}^{3/2} drd\theta =\frac{62}{5}\mathrm{\pi }$

Step 1: Draw the region.

The region determined by the limits is shown below, 

Step 2: Evaluate the integral

$$\begin{gathered} I={\int }_{\mathrm{\pi }/2}^{3\mathrm{\pi }/2}{\int }_1^4{r}^{3/2} drd\theta \\ I={\int }_{\mathrm{\pi }/2}^{3\mathrm{\pi }/2}{\left[\frac{r^{{\displaystyle \frac{3}{2}}+1}}{{\displaystyle \frac{3}{2}}+1}\right]}_1^4 d\theta \\ I=\frac{2}{5}{\int }_{\mathrm{\pi }/2}^{3\mathrm{\pi }/2}{\left[r^{\frac{5}{2}}\right]}_1^4 d\theta \\ I=\frac{2}{5}{\int }_{\mathrm{\pi }/2}^{3\mathrm{\pi }/2}\left[4^{\frac{5}{2}}-1^{\frac{5}{2}}\right]d\theta \\ I=\frac{2\times 31}{5}{\int }_{\mathrm{\pi }/2}^{3\mathrm{\pi }/2}d\theta \\ I=\frac{62}{5}\left[\frac{3\mathrm{\pi }}{2}-\frac{\mathrm{\pi }}{2}\right] \\ I=\frac{62}{5}\mathrm{\pi } \end{gathered}$$